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Proof step 20 from proof middle-coef-larger-strongest (previous step)

0 ≤ 2 * i, 4 ^ i = 1 + Σ countFun(1, 2 * i - 1, λ j. binom_coef(2 * i, j)) + 1, binom_coef(2 * i, i) * 2 * i < 4 ^ i, 0 ≤ i, 0 < i, ∀ i0. i0 = binom_coef(2 * i, 0) ∨ i0 ∈ countFun(1, 2 * i - 1, λ j. binom_coef(2 * i, j)) ∨ i0 = 1 → i0 ≤ binom_coef(2 * i, i) ⊦ binom_coef(2 * i, i) ≠ binom_coef(2 * i, 0) ∧ ¬ binom_coef(2 * i, i) ∈ countFun(1, 2 * i - 1, λ j. binom_coef(2 * i, j)) ∧ binom_coef(2 * i, i) ≠ 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: middle-coef-not-one: ⊦ 0 < i → binom_coef(2 * i, i) ≠ 1;%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%zero-ge: ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); from specification int-mult%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2